A solution has a [tex]$\left[ H _3 O ^{+}\right]$[/tex] of [tex]$1 \times 10^{-5} M$[/tex]. What is the [tex]$\left[ OH ^{-}\right]$[/tex] of the solution?

A. [tex]$9 M$[/tex]
B. [tex]$14 M$[/tex]
C. [tex]$1 \times 10^{-9} M$[/tex]
D. [tex]$1 \times 10^{-14} M$[/tex]



Answer :

To determine the concentration of [tex]\([ \text{OH}^- ]\)[/tex] in a solution given the concentration of [tex]\([ \text{H}_3 \text{O}^+ ]\)[/tex], we need to use the water dissociation constant, [tex]\(K_w\)[/tex].

At 25°C, the value of [tex]\(K_w\)[/tex] is [tex]\(1.0 \times 10^{-14}\)[/tex]. This constant represents the product of the molar concentrations of hydrogen ions [tex]\([ \text{H}_3 \text{O}^+ ]\)[/tex] and hydroxide ions [tex]\([ \text{OH}^- ]\)[/tex] in water:
[tex]\[ K_w = [ \text{H}_3 \text{O}^+ ] \times [ \text{OH}^- ] \][/tex]

Given:
[tex]\[ [ \text{H}_3 \text{O}^+ ] = 1.0 \times 10^{-5} \, \text{M} \][/tex]

We need to find:
[tex]\[ [ \text{OH}^- ] \][/tex]

Rearrange the equation:
[tex]\[ [ \text{OH}^- ] = \frac{K_w}{ [ \text{H}_3 \text{O}^+ ] } \][/tex]

Substitute the known values:
[tex]\[ [ \text{OH}^- ] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-5}} \][/tex]

Perform the division:
[tex]\[ [ \text{OH}^- ] = 1.0 \times 10^{-14} \div 1.0 \times 10^{-5} \][/tex]

Since we are dividing powers of 10, we subtract the exponents:
[tex]\[ [ \text{OH}^- ] = 1.0 \times 10^{(-14) - (-5)} \][/tex]
[tex]\[ [ \text{OH}^- ] = 1.0 \times 10^{-9} \][/tex]

Thus, the concentration of [tex]\([ \text{OH}^- ]\)[/tex] in the solution is:
[tex]\[ 1.0 \times 10^{-9} \, \text{M} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{1 \times 10^{-9} \, \text{M}} \][/tex]